Problem: Simplify and expand the following expression: $ \dfrac{5}{3r - 6}- \dfrac{2}{r - 6}+ \dfrac{1}{r^2 - 8r + 12} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the first term: $ \dfrac{5}{3r - 6} = \dfrac{5}{3(r - 2)}$ We can factor the quadratic in the third term: $ \dfrac{1}{r^2 - 8r + 12} = \dfrac{1}{(r - 2)(r - 6)}$ Now we have: $ \dfrac{5}{3(r - 2)}- \dfrac{2}{r - 6}+ \dfrac{1}{(r - 2)(r - 6)} $ The least common multiple of the denominators is: $ 3(r - 2)(r - 6)$ In order to get the first term over $3(r - 2)(r - 6)$ , multiply by $\dfrac{r - 6}{r - 6}$ $ \dfrac{5}{3(r - 2)} \times \dfrac{r - 6}{r - 6} = \dfrac{5(r - 6)}{3(r - 2)(r - 6)} $ In order to get the second term over $3(r - 2)(r - 6)$ , multiply by $\dfrac{3(r - 2)}{3(r - 2)}$ $ \dfrac{2}{r - 6} \times \dfrac{3(r - 2)}{3(r - 2)} = \dfrac{6(r - 2)}{3(r - 2)(r - 6)} $ In order to get the third term over $3(r - 2)(r - 6)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{1}{(r - 2)(r - 6)} \times \dfrac{3}{3} = \dfrac{3}{3(r - 2)(r - 6)} $ Now we have: $ \dfrac{5(r - 6)}{3(r - 2)(r - 6)} - \dfrac{6(r - 2)}{3(r - 2)(r - 6)} + \dfrac{3}{3(r - 2)(r - 6)} $ $ = \dfrac{ 5(r - 6) - 6(r - 2) + 3} {3(r - 2)(r - 6)} $ Expand: $ = \dfrac{5r - 30 - 6r + 12 + 3}{3r^2 - 24r + 36} $ $ = \dfrac{-r - 15}{3r^2 - 24r + 36}$